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The enthalpy of formation of ammonia is ...

The enthalpy of formation of ammonia is `-46.0 KJ mol^(-1)` . The enthalpy change for the reaction `2NH_(3)(g)rarr N_(2)(g)+3H_(2)(g)` is :

A

`+184 kJ`

B

`+23 kJ`

C

`+92 kJ`

D

`+46 kJ`

Text Solution

Verified by Experts

The correct Answer is:
C
`2NH_(3)(g) to N_(2)(g) + 3H_(2)(g)`
`DeltaH_(r )= - (2 xx "enthalpy of formation of" NH_3) = -(2 xx -46) = 92 kJ`.
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